poj 1514 Metal Cutting (dfs+多边形切割)

1514 -- Metal Cutting

　　一道类似于半平面交的题。

　　题意相当简单，给出一块矩形以及最后被切出来的的多边形各个顶点的位置。每次切割必须从一端切到另一端，问切出多边形最少要切多长的距离。

　　因为最短的切割距离肯定是没有多余的切割痕迹的，而且对于多边形的每一条边，都需要至少经过一次，也就是这些是必要切割。又因为最多就只有8条切割的痕迹，所以我们可以枚举每条痕迹的先后次序，然后模拟切割即刻。复杂度O(n!*n)。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

const double EPS = 1e-8;
const double PI = acos(-1.0);
template <class T> T sqr(T x) { return x * x;}
struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
} ;
typedef Point Vec;
Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);}
Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);}
Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);}
Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);}
inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
bool operator < (Point a, Point b) { return sgn(a.x - b.x) < 0 || sgn(a.x - b.x) == 0 && a.y < b.y;}
bool operator == (Point a, Point b) { return sgn(a.x - b.x) == 0 && sgn(a.y - b.y) == 0;}

inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}
inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}
inline double dotDet(Point o, Point a, Point b) { return dotDet(a - o, b - o);}
inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}
inline double vecLen(Vec x) { return sqrt(dotDet(x, x));}
inline Vec vecUnit(Vec x) { return x / vecLen(x);}
inline Vec normal(Vec x) { return Vec(-x.y, x.x) / vecLen(x);}
inline bool onSeg(Point x, Point a, Point b) { return sgn(crossDet(x, a, b)) == 0 && sgn(dotDet(x, a, b)) < 0;}

int segIntersect(Point a, Point c, Point b, Point d) {
    Vec v1 = b - a, v2 = c - b, v3 = d - c, v4 = a - d;
    int a_bc = sgn(crossDet(v1, v2));
    int b_cd = sgn(crossDet(v2, v3));
    int c_da = sgn(crossDet(v3, v4));
    int d_ab = sgn(crossDet(v4, v1));
//    cout << a_bc << ' ' << b_cd << ' ' << c_da << ' ' << d_ab << endl;
    if (a_bc * c_da > 0 && b_cd * d_ab > 0) return 1;
    if (onSeg(b, a, c) && c_da) return 2;
    if (onSeg(c, b, d) && d_ab) return 2;
    if (onSeg(d, c, a) && a_bc) return 2;
    if (onSeg(a, d, b) && b_cd) return 2;
    return 0;
}

Point lineIntersect(Point P, Vec v, Point Q, Vec w) {
    Vec u = P - Q;
    double t = crossDet(w, u) / crossDet(v, w);
    return P + v * t;
}

struct Poly {
    vector<Point> pt;
    Poly() { pt.clear();}
    ~Poly() {}
    Poly(vector<Point> &pt) : pt(pt) {}
    Point operator [] (int x) const { return pt[x];}
    int size() { return pt.size();}
} ;

Poly cutPoly(Poly &poly, Point a, Point b) {
    Poly ret = Poly();
    int n = poly.size();
    for (int i = 0; i < n; i++) {
        Point c = poly[i], d = poly[(i + 1) % n];
        if (sgn(crossDet(a, b, c)) >= 0) ret.pt.push_back(c);
        if (sgn(crossDet(b - a, c - d)) != 0) {
            Point ip = lineIntersect(a, b - a, c, d - c);
            if (onSeg(ip, c, d)) ret.pt.push_back(ip);
        }
    }
    return ret;
}

const double dir[4][2] = { {0.0, 0.0}, {1.0, 0.0}, {1.0, 1.0}, {0.0, 1.0}};
const double FINF = 1e100;
bool vis[10];
Point rec[10];
double minLen;

void dfs(int p, int n, Poly &poly, double len) {
//    for (int i = 0; i < n; i++) cout << vis[i]; cout << endl;
    if (p >= n) {
        minLen = min(len, minLen);
        return ;
    }
    for (int i = 0; i < n; i++) {
//        cout << i << endl;
        if (vis[i]) continue;
        vis[i] = true;
        Vec d = vecUnit(rec[(i + 1) % n] - rec[i]) * 500.0;
//        cout << rec[(i + 1) % n].x << ' ' << rec[(i + 1) % n].y << " !!! " << rec[i].x << ' ' << rec[i].y << endl;
//        cout << d.x << ' ' << d.y << endl;
        Point s = rec[(i + 1) % n] + d, t = rec[i] - d;
        vector<Point> ip;
        ip.clear();
        for (int j = 0, sz = poly.size(); j < sz; j++) {
            if (segIntersect(s, t, poly[j], poly[(j + 1) % sz])) {
                ip.push_back(lineIntersect(s, t - s, poly[j], poly[(j + 1) % sz] - poly[j]));
//                cout << "has one" << endl;
            }
        }
        sort(ip.begin(), ip.end());
        int cnt = (int) (unique(ip.begin(), ip.end()) - ip.begin());
        if (cnt != 2) {
            puts("shit!!");
            while (1) {}
        }
        Poly tmp = cutPoly(poly, s, t);
        dfs(p + 1, n, tmp, len + vecLen(ip[0] - ip[1]));
        vis[i] = false;
    }
}

int main() {
//    freopen("in", "r", stdin);
    double x, y;
    int n;
    while (cin >> x >> y) {
        Poly tmp = Poly();
        for (int i = 0; i < 4; i++) tmp.pt.push_back(Point(dir[i][0] * x, dir[i][1] * y));
        memset(vis, false, sizeof(vis));
        cin >> n;
        for (int i = 0; i < n; i++) cin >> rec[i].x >> rec[i].y;
        minLen = FINF;
        dfs(0, n, tmp, 0.0);
        printf("Minimum total length = %.3f
", minLen);
    }
    return 0;
}

——written by Lyon